Given the curves $x^2=16y$ and $4y^2=x$
(i) Calculate the area of the region bounded by the two curves.
(ii) Find the volume of the solid formed when this region is rotated about the x-axis.
(iii) Find the volume of the solid formed when this region is rotated about the y-axis.
I'm not sure where to start with this question. I know I have to integrate to find the area and have sketched the curves. They intersect at (0,0) and (4,1). The volume I have no idea how to integrate using the x and y-axis. Help and working would be appreciated.
(i)
To find the area bounded by two curves $f(x)$and $g(x)$ where $f(x) \geq g(x)$ over the domain of integration $[a, b]$, we simply calculate the integral $\int_a^b f(x) - g(x) dx$. Things can start to get a bit more involved when $\int_a^b f(x) dx \lt 0$ or $\int_a^b g(x) dx \lt 0$, but if you draw the graph and investigate the regions $\int_a^b f(x) dx$ and $\int_a^b g(x) dx$ as well as their difference, the formula becomes clear. For questions where one of $\int_a^b f(x) dx \lt 0$ or $\int_a^b g(x) dx \lt 0$, you can easily cross-check to make sure you get the signs of the integrals correct.
For this question, then, we have:
$A = \int_0^4 \frac{x^{\frac{1}{2}}}{2} - \frac{x^2}{16} dx$
$ = \frac{1}{2} \int_0^4 x^{\frac{1}{2}} - \frac{x^2}{8} dx$
$ = \frac{1}{2}[\frac{2x^{\frac{3}{2}}}{3} - \frac{x^3}{24}]_0^4$
$ = \frac{1}{2}[(\frac{16}{3} - \frac{64}{24}) - (0 - 0)]$
$ = \frac{4}{3}$
(ii)
Similarly to before, the volume of the total solid is the volume of the larger solid minus the volume of the smaller solid, since the smaller solid is entirely contained within the larger one. The integral is thus:
$V_x = \pi \int_0^4 (\frac{x^\frac{1}{2}}{2})^2 dx - \pi \int_0^4 (\frac{x^2}{16})^2 dx$
$ = \pi \int_0^4 \frac{x}{4} - \frac{x^4}{16^2} dx$
$ = \pi [\frac{x^2}{8} - \frac{x^5}{5 \times 16^2}]$
$ = \pi [(2 - \frac{4^5}{5\times 16^2}) - (0 - 0)]$
$ = 2\pi - \frac{4\pi}{5}$
$ = \frac{6\pi}{5}$
(iii)
We can do the same thing for the $y$ axis, transforming our functions so they are functions of $y$ instead of $x$, and making sure to change our bounds of integration. Also note now that the larger solid is the one formed by rotating $x^2 = 16y$ around the $y$ axis, where in the previous example it was the one formed by rotating $\frac{x^\frac{1}{2}}{2}$ around the $x$ axis.
$V_y = \pi \int_0^1 (4 y^\frac{1}{2})^2 dy - \pi \int_0^1 (4y^2)^2 dy$
$ = \pi \int_0^1 16y - 16y^4 dy$
$ = 16\pi [\frac{y^2}{2} -\frac{y^5}{5}]$
$ = 16\pi [(\frac{1}{2} - \frac{1}{5}) - (0 - 0)]$
$ = \frac{24\pi}{5}$